Origins act oddly over complex numbers. Offered this, just how do non-integer powers act over adverse numbers? Extra especially:

• Can we specify fractional powers such as $(-2)^{-1.5}$?
• Can we specify illogical powers $(-2)^\pi$?
0
2019-05-06 02:07:43
Source Share

We can specify intricate powers as

$z^w := e^{(l(z)*w)}$ for $z,w \in \mathbb{C}$ and also an intricate logarithm function $l$

certainly you require to see to it there in fact exists such a logarithm $l:U\to\mathbb{C}$ function ($U$ requires to be a merely - linked part of $\mathbb{C}\setminus\{0\}$

0
2019-05-08 13:53:27
Source

Wikipedia claims "Complex powers of favorable reals are specified using e ^ x as in area Complex powers of favorable actual numbers over. These are continual features. Attempting to expand these features to the basic instance of noninteger powers of complex numbers that are negative reals brings about troubles. Either we specify alternate features or multivalued features."

So allow me claim a couple of even more words. Basically what we are seeking is $z^a=e^{a log(t)}$. Yet as you might or might not recognize, complex logarithms are bothersome relying on your point of view.

In an answer to Harry's inquiry, I define an even more mindful means to specify the facility rapid function, and also recommend that need to simply inverted this to get your logarithm. In this manner is secure and also will certainly generate your logarithm.

If you are much less unpredictable, and also agree to tackle the branched logarithms, might I recommend $Log(z)=ln\mid z\mid +iArg(z)$ where Arg is the concept argument. This will certainly lead you to branched backers.

0
2019-05-08 13:50:41
Source

Below is my geometric understanding of why illogical powers of downsides are hard to specify. Thus it is possibly not strenuous and also might be incorrect.

The means illogical powers of the actual numbers are generally specified is by restrictions of fractional powers.

For complex numbers, the very same holds true, other than the restricting procedure is extra difficult. We can certainly shore customarily on our actual numbers result and also see that we just require to specify illogical powers on the device circle, given that every various other factors is some favorable actual multiple of a factor on the device circle.

Currently as a whole z - > z ^ n covers the circle itself n times. What does z - > z ^ (1/n) do? Well, it's unclear given that each factor has n feasible factors it can have originated from, specifically if you dividing the circle right into n arcs of size 2pi/n, each of those obtains mapped fully circle. As soon as you pick a beginning arc though, z - > z ^ m maps the beginning arc to various other arcs in the list below means. Dividing your beginning arc of size 2pi/n right into arcs of size 2pi/ (nm) and afterwards each of the little arcs obtains mapped to a large arc that is 2pim/n far from the previous large arc.

The factor you pick arcs in contrast to strangely dispersed distinct collections, is due to the fact that you desire exponentiation to be continual and also therefore you desire the inverted photo collections to be as the entire circle is attached.

There is no worry with fractional powers of - 1, you have n selections for beginning arc. Yet if you desire your exponentiation to have some form of connection relative to the backer, after that you need to be picking branches (arcs) for z - > z ^ (1/n) that cohere, i.e. such that for large n the origins get closer and also closer with each other. This is done by calling for that the umpteenth origin of 1 is constantly 1, that makes every one of the arcs right into areas of 1. Yet this suggests that all the factors with argument in between 0 and also pi get mapped in the fifty percent - arc over 1, and also all the factors with argument in between pi and also 2pi get mapped in the fifty percent - arc listed below 1.

Therefore if you come close to - 1 from above and also from below, both restrictions of the umpteenth origin will certainly be various, and also therefore you can not have continual exponentiation at - 1.

Therefore (you can do some even more visualization if you desire), the illogical powers of - 1 can not also be specified as restrictions of sensible powers.

0
2019-05-08 12:28:42
Source

As various other posters have actually shown, the trouble is that the facility logarithm isn't well - specified on $\mathbb{C}$. This relates to my remarks in a recent question concerning the square origin not being well - specified (given that certainly $\sqrt{z} = e^{ \frac{\log z}{2} }$).

One perspective is that the facility rapid $e^z : \mathbb{C} \to \mathbb{C}$ does not actually have domain name $\mathbb{C}$. As a result of periodicity it actually has domain name $\mathbb{C}/2\pi i \mathbb{Z}$. So one means to specify the facility logarithm is not as a function with array $\mathbb{C}$, yet as a function with array $\mathbb{C}/2\pi i \mathbb{Z}$. Hence as an example $\log 1 = 0, 2 \pi i, - 2 \pi i, ...$ etc.

So what are we doing when we do not do this? Well, allow us intend that for the time being we have actually determined that $\log 1 = 0$. This is just how we get various other values of the logarithm : making use of power collection, we can specify $\log (1 + z)$ for any kind of $z$ with $|z| < 1$. We can currently select any kind of number in this circle and also take a power collection development concerning that number to get a various power collection whose circle of merging is elsewhere. And also by repetitively transforming the facility of our power collection, we can calculate various values of the logarithm. This is called analytic extension , and also commonly it continues by picking a (claim, smooth) course from $1$ to a few other intricate number and also taking power collection around various factors because course.

The trouble you promptly face is that the value of $\log z$ relies on the selection of course from $1$ to $z$. As an example, the course $z = e^{2 \pi i t}, 0 \le t \le 1$ is a course from $1$ to $1$, and also if you analytically proceed the logarithm on it you will certainly get $\log 1 = 2 \pi i$. Which is not what you desired. (This is basically the like the shape indispensable of $\frac{1}{z}$ along this shape.)

One means around this trouble is to randomly pick a ray from the beginning and also proclaim that you are not permitted to analytically proceed the logarithm via this ray. This is called picking a branch cut, and also it is not approved, so I do not like it.

There is an additional means to settle this scenario, which is to take into consideration the Riemann surface $(z, e^z) \subset \mathbb{C}^2$ and also to consider the logarithm as the estimate to the first coordinate from this surface area to $\mathbb{C}$. So all the troubles we have actually run into over have actually resulted from the reality that we have actually been attempting to make believe that this estimate has particular buildings that it does not have. A shut course like $z = e^{2\pi i t}$ in which the logarithm begins and also finishes with various values represents a course on this surface area which begins and also finishes at various factors , so there is no opposition. This was Riemann's initial inspiration for specifying Riemann surface areas, and also it is this certain Riemann surface area that powers points like the deposit theory.

0
2019-05-08 12:20:58
Source